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Theorem

If $f$ is a bounded measurable function, then there exists a sequence of simple functions $\{f_{n}\}$ which converge uniformly to $f$.

Proof

Define $f_{n}(x) = \frac{m}{n}$ whenever $\frac{m}{n} \leq f(x) < \frac{m+1}{n}$. We clearly have that $f_{n}(x) \rightarrow f(x)$ (if this isn’t clear, look at what value $f_{n}(x)$ is and in what range the $f(x)$ can be; $f(x)$ essentially gets “squished” between these two values, and $f_{n}(x)$ goes to this value as $n$ increases). In addition, and here is the clever part, notice that for every $x$ it is the case that

$$|f(x) - f_{n}(x)| < \frac{1}{n}$$

which gives us convergence. The fact that $f$ is bounded between, say, $[m,M]$, gives uniformity (why?). $\Box$.

I don't understand why we need $f$ bounded for uniform convergence. Isn't it already enough that $|f(x) - f_{n}(x)| < \frac{1}{n}$ was independent of the $x$ value?

Aseed
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Kashif
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    Your $f_n$ can fail to meet a condition of simple function i.e. being the finite sum of scaled characteristic functions. – BigbearZzz Apr 07 '16 at 15:52

2 Answers2

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Every simple function $f_n$ is bounded, so if $f$ is unbounded, $f-f_n$ is unbounded, and you can't have uniform convergence.

As said in the other answer, your proposed function is, in fact, not a simple function for unbounded $f$

Tryss
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  • A simple function is a finite sum of characteristic functions – Tryss Apr 07 '16 at 15:57
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    I don't get the downvote here? The answer is correct. The range of a simple function is finite. – copper.hat Apr 07 '16 at 15:59
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    @copper.hat The question is "where does this break down for unbounded $f$?", and the answer to that is not that the $f_n$ remain bounded (and thus ruin uniform convergence), but rather that they stop being simple. – Arthur Apr 07 '16 at 16:03
  • The question as I read it is why boundedness of $f$ is needed for uniform convergence. – copper.hat Apr 07 '16 at 16:05
  • @copper.hat I realised that once I read your comment. Alas, by then, five minutes had passed, and I am unable to retract my vote until the answer has been edited. Would it be right of me to do a minor edit in order to be able to retract the downvote, you think? – Arthur Apr 07 '16 at 16:14
  • @Arthur: Of course. – copper.hat Apr 07 '16 at 16:16
  • @Arthur : no problem, I'll do a minor edit – Tryss Apr 07 '16 at 16:16
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I will expand my comment just in case that you don't fully understand it. By definition, a simple function $s$ is a finite sum of scaled characteristic functions of measurable sets, i.e. $s$ is of the form $$ s(x)=\sum_{i=1}^mc_i\chi_{A_i}(x)\ . $$ Let us suppose that $f(x)=\frac 1x$ for concreteness. In your proposed example, $f_n$ would have to be an infinite sum of $c_i\chi_{A_i}$ since $f$ is unbounded. This is the reason why your constructed $f_n$ fails to be a simple function. The boundedness of $f$ is indeed indispensable.

BigbearZzz
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